Question

What are the mean and standard deviation of the probability density function given by `(p(x))/k=3x+1-e^(3x+1)` for `x in [0,1]`, in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Answer 1

The problem asks for the solutions for both the mean and standard deviation in terms of k. Since we are given the interval for the p.d.f. ...

Explanation:

mean `=kint_0^1x(3x+1-e^(3x+1))dx`

mean `=k[x^3+x^2/2+(e^(3x+1)(1-3x))/9]_0^1`

mean `=k(1.5-(2e^4)/9-e/9)~~-10.9350k`

Next, calculate the variance ...

variance `=E(X^2]-[E(X)]^2`

variance `=kint_0^1x^2(3x+1-e^(3x+1))dx-[k(1.5-(2e^4)/9-e/9)]^2`

variance `=k/108 (117+8e-20e^4) - k^2[(4 e^8)/81+(4 e^5)/81-2/3e^4+e^2/81-e/3+2.25]`

variance `~~k(-8.8261)-k^2(119.5732)`

standard deviation`=sqrt("variance")`

standard deviation `~~sqrt(k(-8.8261)-k^2(119.5732)`

This is what the asker wanted ... mean and standard deviation in terms of k.

However, John Garbriel solved for k above. `k~~-0.0676`. So, we can go one step further and derive the approximate numerical value for the mean and s.d.:

mean `~~(-0.0676)(1.5-(2e^4)/9-e/9)~~0.7392`

s.d. `~~sqrt(-0.0676(-8.8261)-(-0.0676^2)(119.5732))~~0.2241`

Hope that helped